Puzzle:

• You are dogsitting for 5 nights.
• Dog will wake you up each night with probability 2/3
• What's the probablity of you being able to sleep for at least 2 full nights without being woken up by the dog?

Solution

• Each night prob of waking up is 2/3
• Each night is an independent event
• 5 nights

P(>=2) = 1 - [P(==0) + P(==1)]

P(==0) means you didn't get even a single night's full sleep. ie, dog woke you up on 5 days.

P(==1) means you were able to sleep well on one day, but the dog woke you up on the other 4 days.

P(==0) = 5C5 * (2/3)⁵

Number of sleepless nights = 5 days
5 days can be chosen in 5C5 ways

--

P(==1) = 5C4 * (2/3)⁴ * (1/3)

Number of sleepless nights = 4 days
5 days can be chosen in 5C4 ways

Therefore,

P(>=2) = 1 - [P(==0) + P(==1)]
       = 1 - [(2/3)⁵ + 5*(1/3)*(2/3)⁴]
       = 0.54

Info

Generally,

• P(A ∩ B) = P(A) + P(B) - P(A ∪ B)
• P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Independent events:

• P(A ∩ B) = P(A) * P(B)
• P(A ∪ B) = P(A) + P(B)

Dependent events:

• P(A ∩ B) = P(A)*P(B/A)
• P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Independent events and disjoint events are not same

• Disjoint events: Events that cannot happen together
  • Eg: A single die showing 1 and 2 at the same time
  • Eg: A coin being head and tail at once
• Independent events: Events occurring independent of each other
  • When 2 coins are flipped together, result of each coin is independent of the other.
  • In a chess game, each move is dependent on the previous moves.

If two events are disjoint, they are dependent unless one of the event is of zero probability.

A coin fip and a die roll simultaneously:

• Coin flip and die roll are independent events
• Result of coin flip has no effect on result of die roll

Example:

  P(coin=H ∧ die=1)  
= P(coin=H)* P(die=1)
=    1/2   *  1/6
=    1/12

---

  P(die=3 ∨ die=5)  
= P(die=3)+ (die=5)  
= 1/6     + 1/6
= 1/3

—

Probability of event union for non-disjoint events:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

P(coin=H) and P(die=1) in our example are not disjoint events.

∵ Coin can be head while die shows 1.

  P(coin=H ∨ die=1)  
= P(coin=H) + P(die=1) - P(coin=H ∧ die=1)
= P(coin=H) + P(die=1) - [P(coin=H) * P(die=1)]
=    (1/2)  + (1/6)    -  (1/2)*(1/6)
=        (4/6)         -  (1/12)
=    7/12

Could also be done like: 

  P(coin=H ∨ die=1)  
= 1 - P(¬(coin=H ∨ die=1))
= 1 - P(¬(coin=H) ∧ ¬(die=1))
= 1 - [P(¬(coin=H)) * P(¬(die=1))]
= 1 - [1 - P(coin=H)] * [1 - P(die=1)]
= 1 - [1 - (1/2)] * [1 - (1/6)]
= 1 - [1/2] * [5/6]
= 1 - [5/12]
= 7/12

P(A ∧ ¬A) = 0

Dependent events.

  P(A ∩ ¬A)
= P(A) * P(¬A/A) 
= P(A) * 0
= 0

—

P(A ∨ ¬A) = 1

Dependent events.

  P(A ∪ ¬A)
= P(A) + P(¬A) - P(A ∩ ¬A)  
= x + (1-x) - 0
= 1

Puzzle

• 6 cars parked in a row
• Prob that a car is taxi (independently) is 1/2
• What is the prob that there are 3 consecutive taxis among the 6 cars?

Solution

Q

Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is.

Cheryl gives them a list of 10 possible dates:

• May 15, May 16, May 19
• Jun 17, Jun 18
• Jul 14, Jul 16
• Aug 14, Aug 15, Aug 17

Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively.

• Albert: I don't know when Cheryl's birthday is, but I know that Bernard doesn't know too.
• Bernard: At first I did't know when Cheryl's birthday is, but I know now.
• Albert: Then I also know when Cheryl's birthday is.

So when is Cheryl's birthday?

A

• A knows month
• B knows day

> Albert: I don't know when Cheryl's birthday is, but I know that Bernard doesn't know too.

Bernad doesn't know means the day repeats in the list of 10 dates.

That means one of:

• Jul 14, Aug 14
• May 15, Aug 15
• May 16, Jul 16
• Jun 17, Aug 17

> Bernard: At first I did't know when Cheryl's birthday is, but I know now.

Q

How do you give change worth £1.5? Use two UK coins. One coin must not be 50p.

For context: UK coins are £2, £1, 50p, 20p, 10p, 5p, 2p, 1p.

A